My question is that: Is $\overline X$ independent of $X_1 - \overline X$, given that $X_1,\ldots,X_n$ are IID random variables?
I was thinking that it is independent, but i don't know the logic behind it.
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They are independent if $X_1$ is normally distributed. Whether they are independent in other situations is another question. – Michael Hardy Sep 30 '17 at 14:29
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An example showing that they are in fact dependent.
if $\bar{X}$ indicates the sample mean then they are not independent. For example take $N=2$ and the $X_i$ to be independent $\operatorname{Bernoulli} (p)$. Then $\bar{X}=(X_1 + X_2)/2$. Suppose we condition on $\bar{X}=0$. Then $X_1 - \bar{X} = 0$. This shows that they are dependent.
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If $X_1,\ldots,X_n\sim\operatorname{i.i.d. N}(\mu,\sigma^2)$ then two linear combinations $a_1X_1+\cdots+a_nX_n$ and $b_1X_1+\cdots+b_nX_n$ are independent if their covariance is $0.$
\begin{align} X_1 & = 1X_1 + 0X_2+0X_3 + \cdots + 0 X_n \\[10pt] X_1-\overline X & = \left(1-\frac 1 n\right) X_1+ \left( \frac{-1} n\right) X_2 + \left( \frac{-1} n\right) X_3 + \left( \frac{-1} n\right) X_4 + \cdots + \left( \frac{-1} n\right) X_n \end{align} Using the fact that covariance is linear in each argument separately, and also that the covariance of independent random variables is $0,$ we can find that $\operatorname{cov}(X_1-\overline X, \overline X) = 0.$